Question

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A sample of drinking water was found to be severely contaminated with chloroform,$\sf CHCl_3$, supposed to be carcinogenic in nature. The level of contamination was 15
ppm (by mass).


(i) Express this in per cent by mass.


(ii) Determine the molality of chloroform in the water sample.​

Answer 1

15 ppm corresponds to 15 g chloroform in 1000,000 g of solution.

(i) Percent by mass =Mass of solutionMass of chloroform×100

Percent by mass =1000,00015×100=1.5×10−3% 

(ii) Molality =Molar mass of chloroform ×(Mass of solution−mass of chloroform)Mass of chloroform×1000 (all masses in g)

Molality =119.5×(1000,000−15)15×1000=1.255×10−4 m.

Answer 2

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15 ppm corresponds to 15 g chloroform in 1000,000 g of solution.

i) Percent by mass =

$\frac{mass \: of \: chloroform}{mass \: of \: solution} \times 100$

Percent by mass =

$\frac{15}{1000000} \times 100$

= 1.5 × 10-³ %

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